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For a 3s-orbital Phi(3s)=(1)/(asqrt(3)...

For a 3s-orbital
`Phi(3s)=(1)/(asqrt(3))((1)/(a_(0)))^(3//2)(6-6sigma+sigma^(2))in^(-sigma//2)`
where `sigma=(2rZ)/(3a_(sigma))`
What is the maximum radial distance of node from nucleus?

A

`((3+sqrt(3))a_(sigma))/(Z)`

B

`(a_(sigma))/(Z)`

C

`(3)/(2)((3+sqrt(3))a_(sigma))/(Z)`

D

`(3a_(sigma))/(Z)`

Text Solution

Verified by Experts

The correct Answer is:
C
`Phi(3s)=0`
`6-6sigma+sigma^(2)=0`
`sigma=((-6)+-sqrt(36-(4(1)6)))/(2xx1)`
`sigma=(6+-sqrt(12))/(2)`
`sigma=(6+-2xxsqrt(3))/(2)`
`sigma=3+sqrt(3)` or `3-sqrt(3)`
`(2rZ)/(3a_(@))=3sqrt(3)`
`r=(3)/(2)(3+sqrt(3))(a_(@))/(Z)`
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